Differential Topology Solution Set #1

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1. Prove that Sn = {x ∈ Rn+1 : |x| = 1} is an n−dimensional manifold.

We make coordinate charts on Sn that cover the whole space. Let U+ = {(x1, . . . , xn+1) ∈Sn such that xn+1 > 0}. We define the diffeomorphism:

φ : U+ −→ V ⊂ Rn

by φ((x1, . . . , xn+1)) = (x1, . . . , xn). Notice that φ is a one-to-one because xn+1 =

1 −√∑n

i=1(xi)2 (and in particular is determined by x1, . . . , xn. φ is clearly infinitely

differentiable, and its inverse is also clearly infinitely differentiable by the samemethods introduced in the book for the case n = 1. Similarly, we define a dif-ferent chart for the Southern hemisphere, where we use the same map on the setU- = {(x1, . . . , xn+1) ∈ Sn such that xn+1 < 0}.

We have now covered the sphere Sn with local charts except on the equator, a copyof Sn-1 given by {(x1, . . . , xn, 0) :

∑i x2i = 1}. Note that it would not be sufficient to

use induction here, since we would not have shown that these points have neigh-borhoods in Sn that are locally diffeomorphic to Rn.

Instead, we note that the first paragraph could have been done for any hemisphere.For any i, let

U+i = {(x1, . . . , xn+1) ∈ Sn : xi > 0}, and U-

i = {(x1, . . . , xn+1) ∈ Sn : xi < 0}

andφi : U+

i −→ V ⊂ Rn

be given by φi((x1, . . . , xn+1)) = (x1, . . . , x̂i, . . . xn+1), where x̂i indicates omittingxi. Then φi is clearly smooth. It is one-to-one (and therefore invertible) because

φ-1i (y1, . . . , yn) = (y1, . . . , yi-1, 1 −

√∑n

j=1(yi)2, yi+1 . . . , yn), where the term in the

middle occurs in the ith position. It is clear that φ-1i is (infinitely) differentiable – to

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check the ith term, one simply checks directly. So φi is a diffeomorphism. Similarly,φi : U-

i → V ⊂ Rn defined the same way is a diffeomorphism. Clearly, any point onSn is in U+

i or U-i for some i.

2. Section 1, #2

3. Section 1, #3

4. Section 1, #5 Show that every k-dimensional vector subspace V of RN is a manifolddiffeomorphic to Rk, and that all linear maps on V are smooth.

Choose any basis {v1, . . . , vk} of V , where vi ∈ RN. Since V is k-dimensional, thereare exactly k vectors in this basis, and the vi are linearly independent. Choose abasis {e1, . . . , ek} of Rk. Let φ(vi) := ei for all i. Since {vi} are linearly independent, φ

extends linearly to a map on all of V , and since the set {ei} is linearly independent,φ : V → Rk is surjective. It is clearly invertible for the same reasoning, whereφ-1(ei) = vi, and φ-1 : Rk → V is the linear extension of the map on the basiselements. Now φ is one-to-one; the fact that it is smooth follows from the nextexercise. Similarly, φ-1 is a linear map that is smooth by the next exercise. Thus φ isa diffeomorphism. The fact that all linear maps on V are smooth also follows fromthe next exercise.

5. Suppose that U is an open set in Rn. Prove that if L : U → Rm is a linear map, thendLx = L for all x ∈ U. Hint. Suppose 0 ∈ U and try to prove the statement first atx = 0.

Let L be a linear map. Then for any h ∈ Rn,

dLx(h) = limt!0

L(x + th) − L(x)

t= limt!0

L(x) + tL(h) − L(x)

t= limt!0

tL(h)

t= L(h).

It follows that dLx = L for any x ∈ U.

6. Section 2, #1

7. Section 2, #4

8. Section 2, #12 A curve in a manifold X is a smooth map t → c(t) of an interval of R1into X. The velocity vector of the curve c at time to, denoted simply dc

dt(t0), is defined

to be the vector dct0(1) ∈ Tx0(X), where x0 = c(t0) and dct0 : R1 → Tx0(X). In thecase that X = Rk and c(t) = (c1(t), . . . , ck(t)) in coordinates, check that

dc

dt(t0) = (c ′1(t0), · · · , c ′k(t0)).

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Prove that every vector in Tx(X) is the velocity vector of some curve in X, and con-versely.

The first step is straightforward: simply take the matrix(dcidt

(t0))

as is the definitionof dct0 . Then we apply it to 1 by doing matrix multiplication:

dc

dt(t0) = dct0(1) = (

dc1

dt(t0), . . . ,

dck

dt(t0)) · 1 =

dc1dt

(t0)

··

dckdt

(t0)

where this last expression is then equal to (c ′1(t0), · · · , c ′k(t0)), as desired.

By construction, we have shown that the velocity vectors of curves in X that gothrough x are indeed elements of Tx(X), which is the converse statement. To provethat every vector in Tx(X) is a velocity vector of a curve, we have to solve the fol-lowing differential equation (on a manifold — however, in this case the “manifold”is just Rk!). Let v ∈ Tx(X); we need to find a curve c(t) such that c(t0) = x anddct0(1) = v. In coordinates, we may write x = (x1, . . . , xk) and v = (v1, . . . , vk).Then we solve individually for the differential equation in one variable: ci(t0) = xi,and c ′i(t0) = vi. By existence of solutions to differential equations, there is a solu-tion to this equation (and initial condition). Let ci(t) be the solution. Then c(t) =

(c1(t), . . . , ck(t)) is the curve that has has velocity vector v.

9. Let f : R2 → R3 be the function

f(x, y) = (x2 − y, x + y, 5).

Write out df in matrix form.

10. Prove that if φ : U → V is a diffeomorphism of open sets in Rn, then

(dφ)-1 = d(φ-1).

Since φ is a diffeomorphism, it is invertible, and φ ◦ φ-1 = id, the identity map.Then by the chain rule,

d(φ ◦ φ-1) = dφ ◦ d(φ-1) = d(id) = id.

Similarly, d(φ-1) ◦ dφ = id. On the other hand, by definition of inverses,

dφ ◦ (dφ)-1 = (dφ)-1 ◦ dφ = id.

The result follows by uniqueness of inverses.

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Additional problems for graduate students, or undergraduate extra credit

11. Section 1, #10 “The” torus is the set of points in R3 at distance b from the circle ofradius a in the xy plane, where 0 < b < a. Prove that these tori are all diffeomorphicto S1 × S1.

To show that two spaces are diffeomorphic, one can give a (global) diffeomorphism.Let Ta;b be the torus described; note that the circle of radius a in the xy plane is de-scribed by Ca = {(x, y, 0) ∈ R3 : x2 + y2 = a2} and Ta;b = {(x, y, z) : d((x, y, z), Ca) =

b} where d is the distance from any point (x, y, z) to the closest point on Ca. This“closest point” is obtained by dropping a perpendicular to the circle Ca. We beginby writing this in polar (NOT spherical) coordinates: let Ca = {(a, θ, 0) : θ ∈ [0, 2π)}.Then Ta;b = {(r, θ, z) : (r − a)2 + z2 = b2}. Meanwhile, we write S1 × S1 in polarcoordinates as well: S1 × S1 = {(1, η1, 1, η2) : ηi ∈ [0, 2π)} ⊂ R2 ⊕ R2.

φ : Ta;b −→ S1 × S1

be given by the following map:

φ((r, θ, z)) = (1, θ, 1, η),

where η is the unique value in [0, 2π) such that b cos(η) = r − a and b sin(η) = z.(Note that I don’t use the inverse trigonometric functions because they restrict theirranges in order to be well-defined functions). It is clear that the image of φ is indeedS1 × S1, and that the map is one-to-one. The inverse map is

φ-1(1, η1, 1, η2) = (r, η1, z)

where r = b cos(η2)+a and z = b sin(η2). It is clear that φ-1 is smooth, since cos andsin are smooth functions. To show that φ is smooth, one can take the derivative of φ

at different points using the appropriate trigonometric function at the appropriatepoint. For example, z > 0, then use

φz>0((r, θ, z)) = (1, θ, 1, arccos(r − a

b))

is a diffeomorphism around these points. Under this (local) diffeomorphism, youalways obtain that η ∈ [0, π), so clearly φz>0 is not a global diffeomorphism. How-ever, this is the restriction of the global φ to the points where z > 0.

12. Section 1, #16

13. Section 2, #9(a-d)

14. Section 2, #10

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